Question

The number ${51}^{49}+{51}^{48}+{51}^{47}+...+51+1$ is divisible by

• A. $10$
• B. $20$
• C. $25$
• D. $50$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $10$
B $50$
C $20$
D $25$

$1+51+{51}^2...+{51}^{49}$
$=\frac{{51}^{50}-1}{51-1}$
$=\frac{{51}^{50}-1}{50}$
Hence it is divisible by $2,5,10,25$ and $50.$
To check if the number is divisible by $20$, we need to check the last two digits of the sum.
We can observe that, the last two digits of the odd powers of $51$ are $51$, while those of the even powers are $01$.
Hence, the last two digits of the sum of the series will be equal to the last two digits of $25(51+01)=1300$ which is $00$.
Hence, the sum is also divisible by $20$.
Hence, all the options are correct.